A dairyman pays $6.4$ per litre of milk. He then adds water in the milk and sells the mixture at $8$ per litre, thereby making a $37.5\%$ profit. What is the proportion of water to milk, in the milk received by the customers?

Selling Price (SP) of the mixture = $8$ per litre.
Profit percentage = $37.5\%=\frac{3}{8}$.

Let the Cost Price (CP) of the mixture per litre be $x$.

$$\text{SP}=\text{CP}\times \left(1+\frac{\text{Profit}\%}{100}\right)$$

$$8=x\times \left(1+0.375\right)=1.375x$$

$$x=\frac{8}{1.375}=\frac{8000}{1375}=\frac{32}{5.5}=\frac{64}{11}$$

Now, using the rule of alligation:
Cost of pure milk = $6.4=\frac{64}{10}$
Cost of water = $0$
Mean cost price of mixture = $\frac{64}{11}$

$$\begin{array}{ccc}\text{Cost of Water}& & \text{Cost of Milk}\\ 0& & \frac{64}{10}\\ & \searrow \swarrow & \\ & \frac{64}{11}& \\ \swarrow & & \searrow \\ \left(\frac{64}{10}-\frac{64}{11}\right)& & \left(\frac{64}{11}-0\right)\end{array}$$

$$\text{Quantity of Water}:\text{Quantity of Milk}=\left(\frac{64}{10}-\frac{64}{11}\right):\frac{64}{11}$$

Divide through by $64$:

$$\left(\frac{1}{10}-\frac{1}{11}\right):\frac{1}{11}$$

$$\frac{11-10}{110}:\frac{1}{11}$$

$$\frac{1}{110}:\frac{1}{11}=\frac{1}{10}:1=1:10$$

Thus, the ratio of water to milk is $1:10$.